Question: What is the extraneous solution to these equations? $\dfrac{x^2 - 5}{x + 7} = \dfrac{2x + 58}{x + 7}$
Explanation: Multiply both sides by $x + 7$ $ \dfrac{x^2 - 5}{x + 7} (x + 7) = \dfrac{2x + 58}{x + 7} (x + 7)$ $ x^2 - 5 = 2x + 58$ Subtract $2x + 58$ from both sides: $ x^2 - 5 - (2x + 58) = 2x + 58 - (2x + 58)$ $ x^2 - 5 - 2x - 58 = 0$ $ x^2 - 63 - 2x = 0$ Factor the expression: $ (x + 7)(x - 9) = 0$ Therefore $x = -7$ or $x = 9$ At $x = -7$ , the denominator of the original expression is 0. Since the expression is undefined at $x = -7$, it is an extraneous solution.